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1 Intersecting chords in the interior of a circle Secants intersecting at a common exterior point. Tangents intersecting at a common exterior point Secant.

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Presentación del tema: "1 Intersecting chords in the interior of a circle Secants intersecting at a common exterior point. Tangents intersecting at a common exterior point Secant."— Transcripción de la presentación:

1 1 Intersecting chords in the interior of a circle Secants intersecting at a common exterior point. Tangents intersecting at a common exterior point Secant and tangent intersecting at a common exterior point FINDING SEGMENTS OF TANGENTS AND SECANTS PROBLEM 1 PROBLEM 2 PROBLEM 3 PROBLEM 4 Standard 21 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2 2 Standard 21: Students prove and solve problems regarding relationships among chords, secants, tangents, inscribed angles, and inscribed and circumscribed polygons of circles. Los estudiantes prueban y resuelven problemas relacionados con cuerdas, secantes, tangentes, ángulos inscritos y polígonos inscritos y circunscritos a círculos. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3 3 E D A B C Statements Reasons a. b. c. d. e. f. If Prove: (AE)(EB)=(CE)(ED) Draw DB AC and Through any 2 points there is a line. Vertical are S Inscribed with a arc are S AA Similarity = Sides of similar S are proportional (AE)(EB)=(CE)(ED) The product of the means is equal to the product of the extremes. AEC BED ACD ABD AEC DEB CE EB AE ED When two chords intersect in a circle, then segment products are equal. Cuando dos cuerdas intersecan en un círculo, entonces el producto de los segmentos es igual. D A B C E Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4 4 3X 5X 75 3 (5X)(3X) = (3)(75) 15X = 225 2 2 15 X =15 2 2 X 3.9 Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5 5 A B q Line q is a SECANT. A secant is a line or segment that intersects a circle in two points. La línea q es una SECANTE. Una secante es una línea o segmento que interseca un círculo en dos puntos. Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6 6 A B C D P If Prove: (PC)(PD)=(PA)(PB) Statements Reasons b. c. d. e. a. f. BCP DAP CPA APC = (PC)(PD)=(PA)(PB) Through any 2 points there is a line Same intercepted arc. AA Similarity Reflexive Property Similar have proportional sides. S The product of the means is equal to the product of the extremes. Draw and to form two that overlap BC S AD PC PA PB PD When two secants intersect a circle and have a common external point the products of the secants and external segments are equal. Cuando dos secantes intersecan un círculo y tienen un punto externo común los productos de las secantes y los segmentos externos son iguales. A B C D P PCBPAD Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7 7 5X 12 10 23 (5X+12) (12) = (10+23) (10) 60X + 144 = 100 + 230 60X + 144 = 330 -144 60X = 186 60 X 3.1 Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8 8 C E D CDA= m 90° AEC= m 90° DC EC Intercepted arc is 180° All radii are HL a. b. c. C A E D (1)Tangents are perpendicular to radii Las tangentes son perpendiculares a los radios (2)Tangents with a common exterior point are Las tangentes con un punto común exterior son d. Reflexive property A CA DCA ECA Standard 21

9 9 12-4Y Y 2 Y 2 = -12+4Y Y +4Y-12= 0 2 +4 -12 Two numbers that multiplied be negative twelve = (+)(-) or (-)(+) Two numbers that added be positive 4=|(+)| >| (-)| (-1)(12) -1+12=11 (-2)(6) -2+6=4 (Y-2) (Y+6) =0 Y-2=0 Y+6=0 +2 -6 Y=2 and Y= -6 Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10 10 A B C D (AB) (BD) (CB) = 2 When a tangent and a secant intersect at a common exterior point, the product of the secant and the external segment equals the square of the tangent. Cuando una tangente y una secante intersecan en un punto común exterior, el producto de la secante y el segmento externo iguala el cuadrado de la tangente. Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11 11 42 30 Y (Y+30) (30) (42) = 2 1764 = 30Y + 900 -900 864 = 30Y 30 Y = 28.8 Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved


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